DESCRIPTION
khkjhTRANSCRIPT
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
SOAL 1
Gunakan program MSA.m untuk untuk menghitung distribusi gaya dalam (geser dan momen).
Lalu, hitung momen serta perpindahan maksimum pada masing-masing bentang pada problem
berikut.
SOAL 2
Lakukan modifikasi pada program MSA.m untuk menyelesaikan problem dengan beda suhu pada
permukaan balok berikut. Lalu, tentukan/gambar distribusi gaya dalam (geser dan momen) dan
hitung momen serta perpindahan maksimum pada masing-masing bentang.
SOAL 3
Lakukan modifikasi pada program MSA.m untuk menyelesaikan problem beban termal pada
rangka batang. Lalu, tentukan perpindahan, gaya batang dan reaksi perletakan dan hitung gayaserta
perpindahan maksimum yang terjadi.
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
(Anda akan diminta mendemostrasikan bahwa modifikasi program bekerja!)
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Jawaban Soal 1 - Tugas 2
Berikut ditampilkan input data beam untuk soal 1 dengan program Matlab :
DataBeam.m
function D=DataBeam
% Units: KN & m
m=3;n=4;
Coord=[0 0;4 0;6 -3;7 (-3-tan(0.389*pi))];Coord(:,3)=0;
Con=[1 2 1 1;2 3 1 1;4 3 1 0];
Re=ones(n,6);Re(:,[1,2,6])=[1 1 1;0 0 0;0 0 0;1 1 0];
Load=zeros(n,6);Load(:,[1,2,6])=[0 0 0;-50 0 -80;0 0 0;0 0 0];
w=[0 -20 0;0 0 0;0 0 0];
E=[200000000 200000000 200000000];
nu=0.3;G=E/(2*(1+nu));
A=[0.06 0.06 1000];
Iz=ones(1,m)*0.0004;
Iy=ones(1,m)*0.0004;
J=ones(1,m)*1;
St=zeros(n,6);
be=zeros(1,m);
D=struct('m',m,'n',n,'Coord',Coord','Con',Con','Re',Re','Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',
J','St',St','be',be');
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Berikut ditampilkan modifikasi program MSA pada Matlab :
MSA.m
function [Q,V,R]=MSA(D)
m=D.m;n=D.n;Ni=zeros(12,12,m);S=zeros(6*n);Pf=S(:,1);Q=zeros(12,m);Qfi=Q;Ei=Q;
for i=1:m
H=D.Con(:,i);C=D.Coord(:,H(2))-D.Coord(:,H(1));e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];c=D.be(i);
[a,b,L]=cart2sph(C(1),C(3),C(2));ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);
r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];T=kron(eye(4),r);
x=D.A(i)/L;g=D.G(i)*D.J(i)/(D.E(i)*L);ez=D.E(i)*D.Iz(i)/(D.A(i)*D.G(i));ey=D.E(i)*D.Iy(i)/(D.A(i)*D.G
(i));
z=D.Iz(i)/(L*(L^2/12+ez))*[1 L/2 (L^2/3+ez) (L^2/6-ez)];
y=D.Iy(i)/(L*(L^2/12+ey))*[1 L/2 (L^2/3+ey) (L^2/6-ey)];
K1=diag([x,z(1),y(1)]);K2=[0 0 0;0 0 z(2);0 -y(2) 0];K3=diag([g,y(3),z(3)]);K4=diag([-g,y(4),z(4)]);
K=D.E(i)*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];
w=D.w(:,i)';Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';Qfs=K*T*D.St(e)';
A=diag([0 -0.5 -0.5]);B(2,3)=1.5/L;B(3,2)=-1.5/L;W=diag([1,0,0]);Z=zeros(3);M=eye(12);p=4:6;q=10:12;
switch 2*H(3)+H(4)
case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];
end
K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);Qfi(:,i)=M*Qf;Pf(e)=Pf(e)+T'*M*(Qf+Qfs);Ei(:,i)=e;
end
V=1-(D.Re|D.St);f=find(V);V(f)=S(f,f)\(D.Load(f)-Pf(f));R=reshape(S*V(:)+Pf,6,n);R(f)=0;V=V+D.St;
for i=1:m
Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i);
end
clc
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Berikut adalah output yang dihasilkan dari program tersebut :
Tabel Internal Forces
Tabel Defleksi
Elemen 1 Elemen 2 Elemen 3
Fx beginning 65.95341 45.35912 46.68942
Fy beginning 36.12072 -11.0658 -2.22E-15
Fz beginning 0 0 2.74E-22
Mx beginning 0 0 -1.04E-12
My beginning 0 0 -8.01E-22
Mz beginning 24.58457 -39.8983 -8.88E-16
Fx end -65.9534 -45.3591 -46.6894
Fy end 43.87928 11.06581 2.22E-15
Fz end 0 0 -2.74E-22
Mx end 0 0 1.04E-12
My end 0 0 0
Mz end -40.1017 0 0
Support 1 Support 2 Support 3 Support 4
Dx 0 -2.20E-05 0.000917 0
Dy 0 -0.00031 0.000333 0
Dz 0 0 0 0
Rx 0 0 0 0
Ry 0 0 0 0
Rz 0 -0.00028 0.000615 -0.00033
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Tabel Reaksi Perletakan
Tabel Output Gaya Dalam
Support 1 Support 2 Support 3 Support 4
Fx support 65.95341 0 0 -15.9534
Fy support 36.12072 0 0 43.87928
Fz support 0 0 -1.95E-15 1.95E-15
Mx support 0 0 -3.57E-13 3.57E-13
My support 0 0 9.81E-13 -9.81E-13
Mz support 24.58457 0 0 0
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Jawaban Soal 2 Tugas 2
Berikut ditampilkan input data beam untuk soal 1 dengan program Matlab :
DataBeam.m
function D=DataBeam
m=2;n=3;
Coord=[0 0;4 0;8 0];Coord(:,3)=0;
Con=[1 2 1 1;2 3 1 1];
Re=ones(n,6);Re(:,[1,2,6])=[1 1 1;0 0 0;1 1 1];
Load=zeros(n,6);
w=zeros(m,3);
E=[400000000 200000000];
nu=0.30;
G=E/(2*(1+nu));
A=ones(1,m)*0.02;
Iz=ones(1,m)*50e-6;
Iy=ones(1,m)*50e-6;
J=ones(1,m)*1000;
St=zeros(n,6);
be=zeros(1,m);
TTop=[40 0];
TBot=[25 0];
Tr=28;
Diameter=[0.182 0];
Alpha=12e-6;
D=struct('Diameter',Diameter','Tr',Tr,'TBot',TBot','TTop',TTop','Alpha',Alpha,'m',m,'n',n,'Coord',Coord','Co
n',Con','Re'Re','Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',J','St',St','be',be');
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Berikut ditampilkan modifikasi program MSA pada Matlab :
MSA.m function [Q,V,R]=MSA(D)
m=D.m;
n=D.n;
Ni=zeros(12,12,m);
S=zeros(6*n);
Pf=S(:,1);
Q=zeros(12,m);
Qfi=Q;
Ei=Q;
for i=1:m
H=D.Con(:,i);
C=D.Coord(:,H(2))-D.Coord(:,H(1));
e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];
c=D.be(i);
[a,b,L]=cart2sph(C(1),C(3),C(2));
ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);
r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];
T=kron(eye(4),r);
co=2*L*[6/L 3 2*L L];
x=D.A(i)*L^2;
y=D.Iy(i)*co;
z=D.Iz(i)*co;
g=D.G(i)*D.J(i)*L^2/D.E(i);
K1=diag([x,z(1),y(1)]);
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
K2=[0 0 0;0 0 z(2);0 -y(2) 0];
K3=diag([g,y(3),z(3)]);
K4=diag([-g,y(4),z(4)]);
K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];
w=D.w(:,i)';
Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';
Mt=(D.E(i,1)*D.Iz(i,1)*D.Alpha*(D.TTop(i,1)-D.TBot(i,1))/D.Diameter(i,1));
Pt=(D.E(i,1)*D.A(i,1)*D.Alpha*((0.5*(D.TTop(i,1)+D.TBot(i,1)))-D.Tr));
Pft=zeros(1,m);
Mft=zeros(1,m);
Pft(1,1)=Pt;
Mft(1,1)=Mt;
Qft=[Pft(1,i) 0 0 0 0 -Mft(1,i) -Pft(1,i) 0 0 0 0 Mft(1,i)]';
Qfs=K*T*D.St(e)';
A=diag([0 -0.5 -0.5]);
B(2,3)=1.5/L;B(3,2)=-1.5/L;
W=diag([1,0,0]);
Z=zeros(3);
M=eye(12);
p=4:6;
q=10:12;
switch 2*H(3)+H(4)
case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];end
K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);
Qfi(:,i)=M*(Qf+Qfs+Qft);
Pf(e)=Pf(e)+T'*M*(Qf+Qfs+Qft);
Ei(:,i)=e; end
V=1-(D.Re|D.St);f=find(V);V(f)=S(f,f)\(D.Load(f)-Pf(f));R=reshape(S*V(:)+Pf,6,n);R(f)=0;V=V+D.St;
for i=1:m
Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i); end
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Berikut adalah output yang dihasilkan dari program tersebut :
Tabel Internal Forces
Tabel Defleksi
Tabel Reaksi Perletakan
Elemen 1 Elemen 2
Fx beginning 144 144
Fy beginning -3.5964 -3.5964
Fz beginning 0 0
Mx beginning 0 0
My beginning 0 0
Mz beginning -23.3766 -8.99101
Fx end -144 -144
Fy end 3.596404 3.596404
Fz end 0 0
Mx end 0 0
My end 0 0
Mz end 8.991009 -5.39461
Support 1 Support 2 Support 3
Dx 0 0.000144 0
Dy 0 -0.00048 0
Dz 0 0 0
Rx 0 0 0
Ry 0 0 0
Rz 0 -0.00072 0
Support 1 Support 2 Support 3
Fx support 144 0 -144
Fy support -3.5964 0 3.596404
Fz support 0 0 0
Mx support 0 0 0
My support 0 0 0
Mz support -23.3766 0 -5.39461
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Tabel Output Gaya Dalam
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Jawaban Soal 2 Tugas 3
Berikut ditampilkan input data truss untuk soal 1 dengan program Matlab :
DataF2D.m
function D=DataF2D
% Satuan: kN & m - Beban Thermal
m=3;n=4;
Coord=[0 0;4 0;4 3;0 3];
Coord(:,3)=0;
Con=[1 3 1 0;2 3 1 0;3 4 0 1];
Re=ones(n,6);
Re(:,[1,2,6])=[1 1 0;1 1 0;0 0 0;1 1 0];
Load=zeros(n,6);Load(:,[1,2,6])=[0 0 0;0 0 0;-4 -8 0;0 0 0];
w=zeros(m,3);
E=ones(1,m)*200000000;
nu=0.3;G=E/(2*(1+nu));
A=[4e-5 4e-5 4e-5];
Iz=ones(1,m)*0.0004;
Iy=ones(1,m)*0.0004;
J=ones(1,m)*1000;
St=zeros(n,6);
St(1,2)=-0.0025;
be=zeros(1,m);
T1=zeros(1,m);T1(1,1)=48; %T1 pada member 1
T2=zeros(1,m);T2(1,1)=48; % T2 pada member 1
d=ones(1,m);d(1,1)=1; % d pada member 1
Tr=28; % suhu ruangan
alfa=12e-6; % koefisien muai material
D=struct('Tr',Tr,'d',d','T2',T2','T1',T1','alfa',alfa,'m',m,'n',n,'Coord',Coord','Con',Con','Re',Re','Load',Load',
'w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',J','St',St','be',be');
Berikut ditampilkan modifikasi program MSA pada Matlab :
MSA.m function [Q,V,R]=MSA(D)
m=D.m;
n=D.n;
Ni=zeros(12,12,m);
S=zeros(6*n);
Pf=S(:,1);
Q=zeros(12,m);
Qfi=Q;
Ei=Q;
for i=1:m
H=D.Con(:,i);
C=D.Coord(:,H(2))-D.Coord(:,H(1));
e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];
c=D.be(i);
[a,b,L]=cart2sph(C(1),C(3),C(2));
ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);
r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];
T=kron(eye(4),r);
co=2*L*[6/L 3 2*L L];
x=D.A(i)*L^2;
y=D.Iy(i)*co;
z=D.Iz(i)*co;
g=D.G(i)*D.J(i)*L^2/D.E(i);
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
K1=diag([x,z(1),y(1)]);
K2=[0 0 0;0 0 z(2);0 -y(2) 0];
K3=diag([g,y(3),z(3)]);
K4=diag([-g,y(4),z(4)]);
K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];
w=D.w(:,i)';
Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';
Mt=(D.E(i,1)*D.Iz(i,1)*D.Alpha*(D.TTop(i,1)-D.TBot(i,1))/D.Diameter(i,1));
Pt=(D.E(i,1)*D.A(i,1)*D.Alpha*((0.5*(D.TTop(i,1)+D.TBot(i,1)))-D.Tr));
Pft=zeros(1,m);
Mft=zeros(1,m);
Pft(1,1)=Pt;
Mft(1,1)=Mt;
Qft=[Pft(1,i) 0 0 0 0 -Mft(1,i) -Pft(1,i) 0 0 0 0 Mft(1,i)]';
Qfs=K*T*D.St(e)';
A=diag([0 -0.5 -0.5]);
B(2,3)=1.5/L;B(3,2)=-1.5/L;
W=diag([1,0,0]);
Z=zeros(3);
M=eye(12);
p=4:6;
q=10:12;
switch 2*H(3)+H(4)
case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];end
K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);
Qfi(:,i)=M*(Qf+Qfs+Qft);
Pf(e)=Pf(e)+T'*M*(Qf+Qfs+Qft);
Ei(:,i)=e; end
V=1-(D.Re|D.St);f=find(V);V(f)=S(f,f)\(D.Load(f)-Pf(f));R=reshape(S*V(:)+Pf,6,n);R(f)=0;V=V+D.St;
for i=1:m
Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i); end
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Tugas Analisis Struktur II Kevin Nathaniel / 15012109
Berikut adalah output yang ditampilkan dari program tersebut :
Tabel Internal Forces and Moments
Tabel Defleksi
Tabel Reaksi Perletakan
Elemen 1 Elemen 2 Elemen 3
Fx beginning 0.47037 6.277778 1.703704
Fy beginning -3.84 3.93E-16 1.78E-15
Fz beginning 0 0 3.91E-16
Mx beginning 0 0 8.11E-10
My beginning 0 0 0
Mz beginning -19.2 2.91E-16 0
Fx end -0.47037 -6.27778 -1.7037
Fy end 3.84 -3.93E-16 -1.78E-15
Fz end 0 0 -3.91E-16
Mx end 0 0 -8.11E-10
My end 0 0 -1.56E-15
Mz end 0 0 0
Support 1 Support 2 Support 3 Support 4
Dx 0 0 -0.00085 0
Dy -0.0025 0 -0.00235 0
Dz 0 0 0 0
Rx 0 0 0 0
Ry 0 0 0 0
Rz 0.000126 0.000284 -0.00024 -0.00059
Support 1 Support 2 Support 3 Support 4
Fx support 2.296296 -8.76E-18 0 1.703704
Fy support 1.722222 6.277778 0 -1.78E-15
Fz support 0 0 -1.83E-16 1.83E-16
Mx support 0 0 -8.11E-10 8.11E-10
My support 0 0 0 -1.56E-15
Mz support 0 0 0 0